Given: $$ \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B $$ 2.4.1 Use the formula for $$ \cos (A-B) $$ to derive a formula for $$ \sin (A-B) $$. 2.4.2 Prove that $$ \sin 9 A+\sin A=2 \sin 5 A \cdot \cos 4 A $$. 2.4.3 Write down the maximum value of $$ 3^{2 \sin 5 A \cos 4 A} $$

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To derive the formula for $$ \sin (A-B) $$, we can use the identity for $$ \cos (A-B) $$ and the Pythagorean identity $$ \sin^2 x + \cos^2 x = 1 $$.

We know:
$$
\cos (A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B
$$

Using the identity for $$ \sin (A-B) $$:
$$
\sin (A-B) = \sin A \cdot \cos B - \cos A \cdot \sin B
$$

This gives us the derived formula for $$ \sin (A-B) $$. 
We can use the sum-to-product identities to prove this statement. The sum-to-product identity states:
$$
\sin x + \sin y = 2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)
$$

Let $$ x = 9A $$ and $$ y = A $$:
$$
\sin 9A + \sin A = 2 \sin \left( \frac{9A + A}{2} \right) \cos \left( \frac{9A - A}{2} \right)
$$

Calculating the averages:
$$
\frac{9A + A}{2} = 5A \quad \text{and} \quad \frac{9A - A}{2} = 4A
$$

Thus, we have:
$$
\sin 9A + \sin A = 2 \sin 5A \cdot \cos 4A
$$

This proves the statement. 
The expression $$ 2 \sin 5A \cos 4A $$ can be rewritten using the double angle identity:
$$
2 \sin x \cos y = \sin(2x) \quad \text{where } x = 5A \text{ and } y = 4A
$$

The maximum value of $$ \sin(2x) $$ is 1. Therefore, the maximum value of $$ 2 \sin 5A \cos 4A $$ is 1.

Now, substituting this into the expression:
\[
3^{2 \sin 5A \cos 4A} \text{ has a maximum value of } 3^1 = 3.
\
The maximum value of $$ 3^{2 \sin 5A \cos 4A} $$ is $$ 3 $$.

Given:
2.4.1 Use the formula for to derive a formula for .
2.4.2 Prove that .
2.4.3 Write down the maximum value of

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